Суханов Вадим wrote:
[snip code...]
gives:

$VAR1 = {
          'E1' => {
                    'E2' => {
                              'E3' => {
                                        'En' => 1
                                      }
                            },
                    'E3' => {
                              'E1' => {
                                        'Ex' => {
                                                  'Ey' => {
                                                            'Ez' => undef
                                                          }
                                                }
                                      }
                            }
                  }
        };


Ok, but it should be possible to set the same hash, with another array sharing some starting elements and not destroy the previous hash. The result of my previous sample code should be :


$VAR1 = {
          'E1' => {
                    'E2' => {
                              'E3' => {
                                        'En' => 1
                                      }
                            },
                    'Ex' => {
                              'Ey' => {
                                        'Ez' => 1
                                      }
                            }
                  }
        };

        You can't do that with this method...
--
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| |/ \|       Alexandre (Midnite) Jousset      |
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